Questions from Peskin & Schroeder 5.5 about Compton Scattering

  • #1
Wan
3
0

Main Question or Discussion Point

Hi! Just a couple questions on the Compton scattering calculation in P&S. I feel like I'm missing something very simple here but can't quite figure out what it is. On page 166, the amplitude to be evaluated is
$$ i\mathcal M = -ie^2 \epsilon_\mu(k)\epsilon^*_\nu(k^\prime) u_R^\dagger(p^\prime) \sigma^\mu \frac{\overline \sigma \cdot (p-k^\prime)}{-(\omega^2\chi^2+m^2)}\sigma^\nu u_R(p),$$ where ## u_R(p) = \sqrt{2E}(0, 1)^T## and ##u_R(p^\prime) = \sqrt{2E}(1, 0)^T.## P&S argue that the final photon should be right handed otherwise the amplitude vanishes but I don't quite get that. I presumed that if the final photon is right handed then ##\epsilon^\nu = 1/\sqrt{2}(0,1,i,0)^T##, so ##\epsilon_\nu = 1/\sqrt{2}(0,-1,-i,0)^T## and ##\epsilon_\nu^* = (0, -1, i, 0)^T##, then we would have
$$
\sigma^\nu \epsilon_\nu^* = \frac{1}{\sqrt{2}} \bigg[\begin{pmatrix}
0 & 1\\
1 & 0\end{pmatrix} (-1) + \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} i\bigg] = \begin{pmatrix} 0 & 0 \\ -\sqrt{2} & 0 \end{pmatrix},
$$
But then ##\sigma^\nu \epsilon_\nu^*u_R(p) = \begin{pmatrix} 0 & 0 \\ -\sqrt{2} & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 1 \end{pmatrix} = 0##. What did I do wrong?

Also with Figure 5.6 on the next page, I have pretty much the same question as this guy in StackExchange https://physics.stackexchange.com/questions/390121/spin-vs-helicity-conservation, unfortunately no one gave him an answer yet. My understanding is that, before the collision, we have spin/helicity of the photon = 1, spin of electron -1/2; and after the collision we have helicity of the photon = -1, and spin of electron 1/2, therefore one unit of spin is exchanged, is that correct? Any help is appreciated. Thanks!
 

Answers and Replies

  • #2
Wan
3
0
Welp, I wonder why there are no replies. I'll just use this opportunity to ask one more question, anyone knows how the minus sign in the second expression in (5.32) come about? For ## M(e_R^-e_L^+ \rightarrow \mu_L^-\mu_R^+) ## I have it equals to ## \overline v(p^\prime) \gamma^\mu u(p) \cdot \overline u(k) \gamma_\mu v(k^\prime) ## where ## \overline v(p^\prime) \gamma^\mu u(p) = -2E(0,1,i,0) ## (equation (5.29) which describes ##e_R^- e_L^+##) and ## \overline u(k) \gamma_\mu v(k^\prime) = -2E(0,-\cos\theta, -i, \sin\theta) ## (the equation above (5.32) describing ## \mu_L^- \mu_R^+ ##. When I dot them together I got ## M = e^2 (1-\cos\theta)##, without the minus sign. Same thing happens to ## M(e_L^-e_R^+ \rightarrow \mu_R^- \mu_L^+) ##. Where does the minus come from? Many thanks.
 

Related Threads for: Questions from Peskin & Schroeder 5.5 about Compton Scattering

  • Last Post
Replies
17
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
3
Views
869
  • Last Post
Replies
11
Views
3K
Replies
6
Views
3K
Replies
8
Views
301
Replies
8
Views
1K
Replies
3
Views
5K
Top